Решебник по алгебре и начала математического анализа 10 класс Алимов Задание 1179

\[\boxed{\mathbf{1179}\mathbf{.}}\]

\[n = C_{6 + 7}^{2} = C_{13}^{2} = \frac{13!}{(13 - 2)! \bullet 2!} =\]

\[= \frac{13!}{11! \bullet 2} = \frac{13 \bullet 12 \bullet 11!}{11! \bullet 2} =\]

\[= 13 \bullet 6 = 78.\]

\[1)\ оба\ шара\ белого\ цвета:\]

\[m = C_{6}^{2} = \frac{6!}{(6 - 2)! \bullet 2!} = \frac{6!}{4! \bullet 2} =\]

\[= \frac{6 \bullet 5 \bullet 4!}{4! \bullet 2} = 3 \bullet 5 = 15;\]

\[P = \frac{m}{n} = \frac{15}{78} = \frac{5}{26}.\]

\[2)\ оба\ шара\ черного\ цвета:\]

\[m = C_{7}^{2} = \frac{7!}{(7 - 2)! \bullet 2!} = \frac{7!}{5! \bullet 2} =\]

\[= \frac{7 \bullet 6 \bullet 5!}{5! \bullet 2} = 7 \bullet 3 = 21;\]

\[P = \frac{m}{n} = \frac{21}{78} = \frac{7}{26}.\]

\[3)\ один\ шар\ белый,\ другой\ \]

\[черный:\]

\[m = C_{6}^{1} \bullet C_{7}^{1} =\]

\[= \frac{6!}{(6 - 1)! \bullet 1!} \bullet \frac{7!}{(7 - 1)! \bullet 1!} =\]

\[= \frac{6 \bullet 5!}{5!} \bullet \frac{7 \bullet 6!}{6!} = 6 \bullet 7 = 42;\]

\[P = \frac{m}{n} = \frac{42}{78} = \frac{7}{13}.\]

\[4)\ по\ крайней\ мере\ один\ шар\ \]

\[белый:\]

\[A - оба\ шара\ черные;\]

\[\overline{A} - искомое\ событие:\]

\[m = C_{7}^{2} = \frac{7!}{(7 - 2)! \bullet 2!} = \frac{7!}{5! \bullet 2} =\]

\[= \frac{7 \bullet 6 \bullet 5!}{5! \bullet 2} = 7 \bullet 3 = 21;\]

\[P\left( \overline{A} \right) = 1 - P(A) = 1 - \frac{m}{n} =\]

\[= 1 - \frac{21}{78} = 1 - \frac{7}{26} = \frac{19}{26}.\]

\[5)\ по\ крайней\ мере\ один\ шар\ \]

\[черный:\]

\[A - оба\ шара\ белые;\]

\[\overline{A} - искомое\ событие:\]

\[m = C_{6}^{2} = \frac{6!}{(6 - 2)! \bullet 2!} = \frac{6!}{4! \bullet 2} =\]

\[= \frac{6 \bullet 5 \bullet 4!}{4! \bullet 2} = 3 \bullet 5 = 15;\]

\[P\left( \overline{A} \right) = 1 - P(A) = 1 - \frac{m}{n} =\]

\[= 1 - \frac{15}{78} = 1 - \frac{5}{26} = \frac{21}{26}.\]